Given the space of all square-integral functions over $[0,1]$: $L^2([0,1], \mathcal{B}([0,1]), m)$ and a Brownian motion $W_t$ defined on the probability space $(\Omega, \mathcal{F}, P)$, we define $$h(A) = \int_0^1 1_A(t)dW_t$$ for all $A \in \mathcal{B}([0,1])$.
Then for disjoint $(A_n)_{n\geq 1}$, we know that $$h\left(\bigcup\limits_n A_n\right) = \sum_n h(A_n)$$ in $L^2$ and almost surely
It is well known that despite the above identity, $h$ does not define a signed measure pathwisely because although $h\left(\bigcup\limits_n A_n\right) = \sum\limits_n h(A_n)$ almost surely, the set of exceptional $\omega$s depends on the sequence $(A_n)$.
I learned about this fact a long time ago, but I am still a bit confused. The above statement only says that since "the exceptional set of $\omega$s depends on the sequence $A_n$", we didn't manage to prove $h$ is a measure pathwisely. But why is it not possible that "by chance" all the exceptional $\omega$s turns out to be in a negligeable set so we can still define a pathwise measure for almost every $\omega$?
In summary, I am asking for help to prove, no matter how we modify $h(A)$ for negligible $\omega$s, there is always a set $S$ of positive probability, such that for all $\omega \in S$, we can find a sequence of disjoint $A_n \in \mathcal{B}([0,1])$ such that $h\left(\bigcup\limits_n A_n\right)(\omega) \neq \sum\limits_n h(A_n)(\omega)$.
I just realized I can use the fact that for radon measure $\mu$ on $[0,1]$, $f(t) = \mu([0,t])$ is of finite variation.
So in my question, if for $\omega$ in $\Omega_0$ with $P(\Omega_0) = 1$, $h(\cdot)(\omega)$ is a Radon measure on $[0,1]$, then $h([0,t]) = W_t$ is of finite variation. This is contradictory with the fact that Brownian motion has infinite variation almost surely.
Therefore, those $\omega$'s which make $h$ a real measure should be contained in a null set.